3D conformal mappings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:36:39Z http://mathoverflow.net/feeds/question/52209 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52209/3d-conformal-mappings 3D conformal mappings MathGeek 2011-01-16T04:48:24Z 2011-09-03T02:10:57Z <p>Are there analogues to conformal mapping in 3 dimensions? </p> <p>I have a specific example I am trying to solve.. Laplace's equation in 3D with slightly complicated rectilinear boundaries. (Think of solving a harmonic function over a 3D boundary which is a cube but with a sub-cube "bitten" out of one corner.)</p> <p>Laplace's equation is still valid under conformal transformations, so for example in 2D I could take a square domain with a subsquare bitten out of a corner, and apply an inverse tranformation <a href="http://math.fullerton.edu/mathews/c2003/ConformalMapDictionary.3.html" rel="nofollow">like some of these</a> and solve the equation in a simple square domain. </p> <p>Are there similar conformal-like transformations in 3D? Perhaps they wouldn't be called conformal maps, but maybe something exists which would work similarly for my 3D Laplace equations. </p> http://mathoverflow.net/questions/52209/3d-conformal-mappings/52233#52233 Answer by André Henriques for 3D conformal mappings André Henriques 2011-01-16T12:14:39Z 2011-01-16T12:14:39Z <p>I don't know if this is relevant for your question...</p> <p>In <a href="http://arxiv.org/pdf/1005.5464v2" rel="nofollow">http://arxiv.org/pdf/1005.5464v2</a>, the author introduces a notion of "weak conformal map" for 3-dimensional domains, and proves a Riemann mapping theorem for those kinds of maps.</p> <p><b>Definition:</b> given two open subsets $U,V\subset \mathbb R^3$, a smooth map $f:U\to V$ is called weak conformal if, at every $x\in U$, the three eigenvalues of $P_x:\mathbb R^3\to \mathbb R^3$ are in geometric progression. Here, the positive operator $P_x$ is the one coming from the polar decomposition of the tangent map $T_xf:T_x \mathbb R^3 = \mathbb R^3\to T_{f(x)} \mathbb R^3 = \mathbb R^3$.</p>